Class 9

Math

All topics

Circles

In the adjoining figure, chords $AC$ and $BD$ of a circle with centre $O$, intersect at right angles at $E$. If $∠OAB=25_{o}$, calculate $∠EBC$.

We know that $OA$ and $OB$ are the radius

Base angles of an isosceles triangle are equal

So we get

$∠OBA=∠OAB=25_{o}$

Consider $△OAB$

Using the angle sum property

$∠OAB+∠OBA+∠AOB=180_{o}$

By substituting the values

$25_{o}+25_{o}+∠AOB=180_{o}$

On further calculation

$∠AOB=180_{o}−25_{o}−25_{o}$

By subtraction

$∠AOB=1180_{o}−50_{o}$

So we get

$∠AOB=130_{o}$

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference

So we get

$∠AOB=2∠ACB$

It can be written as

$∠ACB=21 ∠AOB$

By substituting the values

$∠ACB=2130 $

By division

$∠ACB=65_{o}$

So we get

$∠ECB=65_{o}$

In the $△BEC$

Using the angle sum property

$∠EBC+∠BEC+∠ECB=180_{o}$

By substituting the values

$∠EBC+90_{o}+65_{o}=180_{o}$

On further calculation

$∠EBC=180_{o}−90_{o}−65_{o}$

By subtraction

$∠EBC=180_{o}−155_{o}$

So we get

$∠EBC=25_{o}$

Therefore , $∠EBC=25_{o}$